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1

Continued fraction (2) can be simplified as $$ \tan\left(\alpha\tan^{-1}z\right)=\cfrac{\alpha z}{1+\cfrac{\frac{(1^2-\alpha^2)z^2}{1\cdot 3}} {1+\cfrac{\frac{(2^2-\alpha^2)z^2}{3\cdot 5}}{1+\cfrac{\frac{(3^2-\alpha^2)z^2}{5\cdot 7}}{1+\ddots}}}}\tag{2a} $$ This is a special case of the following continued fraction due to N$\ddot{\text{o}}$rlund (B.Berndt, ...


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$F(x)$ can be rewritten as $\displaystyle\;\frac{1}{\frac{2}{P(x)} - x}$ where $\displaystyle\;\def\CF{\mathop{\LARGE\mathrm K}} P(x) = \cfrac{1\cdot 2}{1\cdot 2 x + \cfrac{ (1 \cdot 2)(2\cdot 3)}{2\cdot 3 x + \cfrac{(2\cdot 3)(3\cdot 4)}{3\cdot 4 x + \ddots} }} $. The CF $P(x)$ has the form $ \displaystyle\; \CF_{\ell=1}^{\infty} ...


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I think this theorem is discussed in Hardy and Wright. Also Shockley had a very nice discussion, if I recall correctly. The central idea which can be proved fairly readily by mathematical induction, is that if $$\frac{p_n}{q_n}$$ is the $n^{th}$ convergent to the continued fraction of $\sqrt D$, they satisfy a Pell-type equation ...


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Never mind, thanks to the comment by J. M. I found the source of this expression. The series are connected to the exponential integral: $$\text{Ei}(t)=-\int_{-t}^{\infty} \frac{e^{-p}}{p} dp=\gamma+\log |t|+\sum_{n=1}^{\infty} \frac{t^n}{n!n}$$ The continued fraction turns out to be a particular case of incomplete Gamma function: $$\Gamma ...


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For simplicity, we will start with the case $z = 1$. We will assume $k > 0$ and let $\cot\theta = k$ and $\mu_{\pm} = k \pm \sqrt{k^2+1}$. The CF at hand has the form $$ \def\CF{\mathop{\LARGE\mathrm K}} \CF_{\ell=1}^{\infty} \frac{\alpha_\ell\gamma_{\ell-1}}{\beta_\ell} \quad\text{ where }\quad \gamma_0 = 1\quad\text { and }\quad \begin{cases} ...


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(This is a comment that got too long for the comment box.) For fun, I decided to implement this function for complex arguments in Mathematica, and plot its real and imaginary parts. Here's the picture I got: That pole fence jives with the OP's observation that the function is only sensible for $\Re s > -1$. Can it be turned into a series? ...


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The case with $z \neq 1$ can be obtained from the case $z=1$, because you can just factor them out of the infinite fraction while changing $k$ accordingly. Consider $k,m$ fixed, and let $(f_n)$ be the sequence of homographies $f_n(x) = a_n + b_n / x$, with $a_n = (2n-1)km$ and $b_n = ((n-1)m-1)((n+1)m+1)$ so that $\Theta = \lim_{n \to \infty} (m+1)/ f_0 ...


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Your continued fraction is a very special case of the general continued fraction for the quotient of gamma functions conjectured in this post,see corollary (iii) Edited: It is also a special case of the hyperbolic ...


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Using the general solution from this post,we obtain $$x=\dfrac{ik\sqrt[m]{\dfrac{k+i}{k-i}}-\sqrt[m]{\dfrac{k+i}{k-i}}-ik-1}{k\sqrt[m]{\dfrac{k+i}{k-i}}+i\sqrt[m]{\dfrac{k+i}{k-i}}+k-i}$$ $$y=\dfrac{k\sqrt[m]{\dfrac{k-1}{k+1}}-\sqrt[m]{\dfrac{k-1}{k+1}}-k-1}{k\sqrt[m]{\dfrac{k-1}{k+1}}-\sqrt[m]{\dfrac{k-1}{k+1}}+k+1}$$ For $m\gt2$


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Too long for a comment. If you let $a=-1$ and $b=2m+1$ of the general continued fraction in this post, it reduces to the first continued fraction in this post (with $k=1$) and is expressible as a quotient of gamma functions, ...


1

I'll have a stab at this. First, here's a link on how to find the simple continued fraction of a number $x$: http://mathworld.wolfram.com/ContinuedFraction.html And here's a link (see pages 13 - 15) on how to factorize a number $n$, using the simple continued fraction of $x = \sqrt n$: http://wstein.org/edu/2010/414/projects/johnson.pdf From the first ...


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(Too long for a comment.) I. Level $2$ From your other post, we have, $$G_1(x,n)=\cfrac{1}{2x+\cfrac{(-1)(-1+n)} {6x+\cfrac{(1)(1+n)}{10x+\cfrac{(3)(3+n)}{14x+\cfrac{(5)(5+n)}{18x+\ddots}}}}}\tag1$$ with $2v+1 =-1,1,3,5,\dots$ The special case $n=6$, $$G_1(x,\color{brown}6)=\frac{1}{3b}\left(a+\sqrt{c^3}\right)\tag2$$ where, ...


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$\sqrt 2 = (1+1)^{1/2}$ by the binomial theorem: $(1+a)^{1/2}$$ = 1 + (1/2) 1^{-1/2}a-(1/8) 1^{-3/2}a^2+ (3/48) a^3\cdots$ coefficient of the $n^{th}$ term: $c_0 = 1\\c_n = c_{n-1}\frac{(1/2-n)}{n}$ when $n\ge 3, c_n = (-1)^{n+1}\frac{1*3*5*7...(n-2)}{2*4*6*8\cdots n} $ More generally, can you find a series that converges to the alegraic ...


1

Suppose you have a decimal expansion for your irrational number. We'll take $\sqrt2=1.41421\dots$ for example. Then we can write $$\sqrt{2}=1+\frac{4}{10}+\frac{1}{100}+\frac{4}{1000}+\frac{2}{10000}+\frac{1}{100000}+\dots$$ Each term in our sequence is rational (and moreover it is clear how to apply this idea to get an infinite sum of rationals to ...


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Solving the Recurrence Let $$ x = \sqrt 2 + \frac{b}{x} $$ Hence, $$ x = \frac{\sqrt 2 x + b}{x} \implies x^2 = \sqrt 2 x + b \\ x^2 - x\sqrt2 - b = 0 $$ The roots of this equation are $$ x_1, x_2 = \frac{\sqrt2 \pm \sqrt{ \sqrt{2}^2 + 4b}}{2} = \frac{\sqrt{2} \pm \sqrt{2 +4b}}{2} $$ Proving that this works for all Pythogaren triplets Let us ...


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Let $\displaystyle x=\sqrt{2}+\frac{b}{\sqrt{2}+\ldots}$, then \begin{align*} \sqrt{2}+\frac{b}{x} &= x \\ x\sqrt{2}+b &= x^{2} \\ x^{2}-\sqrt{2} \, x-b &= 0 \\ x &= \frac{\sqrt{2}+\sqrt{2+4b}}{2} \\ &= \sqrt{ \left( \frac{\sqrt{2}+\sqrt{2+4b}}{2} \right)^{2} } \\ &= \sqrt{b+1+\sqrt{2b+1}} \\ \end{align*} Take ...


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Just compute your left side: $$x=\sqrt{2}+\frac{b}{x}$$ $$x(x-\sqrt{2})=b$$ $$x^2-\sqrt2 x-b=0$$ $$x=\frac{\sqrt2+ \sqrt{2+4b}}{2}$$ So now you have $$\sqrt2+\sqrt{2+4b}=2\sqrt{a+c}$$ $$2+2+4b+2\sqrt{4+8b}=4(a+c)$$ $$a+c-b-1=\sqrt{1+2b}$$ $$c=1+b-a+\sqrt{1+2b}$$ For that to be a whole number, $b=(x^2-1)/2$ where $x$ is odd. Note that it's valid for number ...


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$x=(1;\overline{3,1})\implies x=1+\cfrac1{3+\cfrac1x}\implies3x^2-3x-1=0$. Thus, $$ (1;\overline{3,1})=\frac{3+\sqrt{21}}6 $$ Therefore, $$ \begin{align} (2;\overline{1,3}) &=2+\frac6{3+\sqrt{21}}\\[3pt] &=\frac{1+\sqrt{21}}2 \end{align} $$ $x=(3;\overline{3})\implies x=3+\cfrac1x\implies x^2-3x-1=0$. Thus, $$ (3;\overline{3})=\frac{3+\sqrt{13}}2 ...


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Quick answer: From link, you can use the formula $$[a,b] = \frac{-ab+\sqrt{ab(ab+4)}}{2a}.$$ Then, you can check the result here with $\frac{1+\sqrt{21}}{2}$



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